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Here is the problem statement (from chapter $3$ of Axler's Linear Algebra Done Right).

Give an example of a vector space $V$ and subspaces $U_1,U_2$ of $V$ such that $U_1 \times U_2$ is isomorphic to $U_1 + U_2$, but $U_1 + U_2$ is not a direct sum.

Hint: the vector space $V$ must be infinite-dimensional.

The only infinite-dimensional vector spaces mentioned in the book up to this point are $\mathbb{F}^{\infty}$ and $P(\mathbb{R})$.

I tried to construct an example using $\mathbb{F}^{\infty}$ by letting $U_1$ be the span of the standard bases $\{e_{2n}\}$ and $U_2$ the span of $\{e_{2n-1}\}$. It seems that at least one of the $U_i$ must be infinite dimensional, or else the example could be constructed with $V$ finite dimensional.

There is an isomorphism between $U_1 \times U_2$ and $U_1 + U_2$, but $U_1 \cap U_2 = \{0\}$ so we have a direct sum, which is what we're trying to avoid. I'm not sure how to proceed from here.

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Modify your example to make the subspaces intersect. For example, we can take $U_1 = \operatorname{span} \{ e_1 \}$ and $U_2 = \operatorname{span} \{ e_i \}_{i \in \mathbb{N}}$. Then $U_1 \cap U_2 = U_1$ so $U_1 + U_2$ is not a direct sum. In addition, $U_1 \times U_2$ is isomorphic to $U_1 + U_2 = U_2$ via the linear map sending

$$ (e_1,0) \mapsto e_1, (0,e_i) \mapsto e_{i + 1}. $$

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  • $\begingroup$ I didn't understand your linear map. However, I believe that it doesn't exist such vector space $V$. Because, if there's it, then the linear map $T$ is injective. But if $U_1 + U_2$ isn't a direct sum, then there's a non-zero $u \in U_1 \cap U_2$. So we have $T(u, -u) = 0$, which implies that T isn't injective. Contradiction. So there isn't such $V$. $\endgroup$ – Rafael Deiga Apr 2 '19 at 11:07
  • $\begingroup$ @RafaelDeiga: It seems that you are assuming that the linear map $T \colon U_1 \times U_2 \rightarrow U_1 + U_2$ must be given by $T(u,v) = u + v$ so that $T(u,-u) = u - u = 0$. However, there's no reason $T$ must have this form. In my example, $e_1 \in U_1 \cap U_2$ but $T(e_1,0) = e_1, T(0,e_1) = e_2$ so $T(e_1,e_1) = T((e_1,0) + (0,e_1)) = T(e_1,0) + T(0,e_1) = e_1 + e_2 \neq 0$. $\endgroup$ – levap Apr 2 '19 at 11:16
  • $\begingroup$ True. Indeed, I have assumed that $T$ is of the form $T(u,v) = u + v$. Now I understood your linear map. $\endgroup$ – Rafael Deiga Apr 2 '19 at 11:21

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